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Transportation Problems

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  1. Transportation Problems • Transportation is considered as a “special case” of LP • Reasons? • it can be formulated using LP technique so is its solution (to p2)

  2. Here, we attempt to firstly define what are them and then studying their solution methods: (to p3)

  3. Transportation Problem • We have seen a sample of transportation problem on slide 29 in lecture 2 • Here, we study its alternative solution method • Consider the following transportation tableau (to p4) (to p6)

  4. Review of Transportation Problem Warehouse supply of televisions sets: Retail store demand for television sets: 1- Cincinnati 300 A - New York 150 2- Atlanta 200 B - Dallas 250 3- Pittsburgh 200 C - Detroit 200 total 700 total 600 (to p5) LP formulation

  5. A Transportation Example (2 of 3) Model Summary and Computer Solution with Excel Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C subject to x1A + x1B+ x1  300 x2A+ x2B + x2C 200 x3A+ x3B + x3C 200 x1A + x2A + x3A = 150 x1B + x2B + x3B = 250 x1C + x2C + x3C= 200 xij 0 Exhibit 4.15 (to p3)

  6. Transportation Tableau • We know how to formulate it using LP technique • Refer to lecture 2 note • Here, we study its solution by firstly attempting to determine its initial tableau • Just like the first simplex tableau! (to p7)

  7. Solution to a transportation problem • Initial tableau • Optimal solution • Important Notes • Tutorials (to p8) (to p23) (to p43) (to p53)

  8. initial tableau • Three different ways: • Northwest corner method • The Minimum cell cost method • Vogel’s approximation method (VAM) • Now, are these initial tableaus given us an Optimal solution? (to p9) (to p12) (to p16) (to p7)

  9. Northwest corner method Steps: • assign largest possible allocation to the cell in the upper left-hand corner of the tableau • Repeat step 1 until all allocations have been assigned • Stop. Initial tableau is obtained Example (to p10)

  10. Northeast corner Step 1 Step 2 Max (150,200) 150 0 0 -- -- 150 -- -- 50 125 -- (to p11) 0 150

  11. Initial tableau of NW corner method • Repeat the above steps, we have the following tableau. • Stop. Since all allocated have been assigned (to p8) Ensure that all columns and rows added up to its respective totals.

  12. The Minimum cell cost method Here, we use the following steps: Steps: Step 1 Find the cell that has the least cost Step 2: Assign as much as allocation to this cell Step 3: Block those cells that cannot be allocated Step 4: Repeat above steps until all allocation have been assigned. Example: (to p13)

  13. Step 1 Find the cell that has the least costStep 2: Assign as much as allocation to this cell Step 1: Step 2: 200 The min cost, so allocate as much resource as possible here Step 3 (to p14)

  14. Step 3: Block those cells that cannot be allocatedStep 4: Repeat above steps until all allocation have been assigned. Second iteration, step 4 Step 3: -- -- (to p15) 75 200 0

  15. The initial solution • Stop. The above tableau is an initial tableau because all allocations have been assigned (to p8)

  16. Vogel’s approximation method Operational steps: Step 1: for each column and row, determine its penalty cost by subtracting their two of their least cost Step 2: select row/column that has the highest penalty cost in step 1 Step 3: assign as much as allocation to the selected row/column that has the least cost Step 4: Block those cells that cannot be further allocated Step 5: Repeat above steps until all allocations have been assigned Example (to p17)

  17. subtracting their two of their least cost Step 1 (8-6) (11-7) (5-4) (to p18) (6-4) (8-5) (11-10)

  18. Steps 2 & 3 Step 2: Highest penalty cost (to p19) Step 3: this has the least cost

  19. Step 4 (to p20) --- ---

  20. Step 5Second Iteration --- (to p21) --- ---

  21. 3rd Iteration of VAM --- --- (to p22) --- ---

  22. Initial tableau for VAM (to p8)

  23. Optimal solution? Initial solution from: Northeast cost, total cost =$5,925 The min cost, total cost =$4,550 VAM, total cost = $5,125 (note: here, we are not saying the second one always better!) It shows that the second one has the min cost, but is it the optimal solution? (to p24)

  24. Solution methods • We need a method, like the simplex method, to check and obtain the optimal solution • Two methods: • Stepping-stone method • Modified distributed method (MODI) (to p25) (to p30) (to p7)

  25. Stepping-stone method Let consider the following initial tableau from the Min Cost algorithm There are Non-basic variables These are basic variables (to p26) Question: How can we introduce a non-basic variable into basic variable?

  26. Introducea non-basic variable into basic variables • Here, we can select any non-basic variable as an entry and then using the “+ and –” steps to form a closed loop as follows: Then we have let consider this non basic variable (to p27)

  27. Stepping stone - + + - (to p28) The above saying that, we add min value of all –ve cells into cell that has “+” sign, and subtracts the same value to the “-ve” cells Thus, max –ve is min (200,25) = 25, and we add 25 to cell A1 and A3, and subtract it from B1 and A3

  28. Stepping stone The above tableaus give min cost = 25*6 + 120*10 + 175*11 175*4 + 100* 5 = $4525 We can repeat this process to all possible non-basic cells in that above tableau until one has the min cost! NOT a Good solution method (to p29)

  29. Getting optimal solution • In such, we introducing the next algorithm called Modified Distribution (MODI) (to p24)

  30. Modified distributed method (MODI) • It is a modified version of stepping stone method • MODI has two important elements: • It determines if a tableau is the optimal one • It tells you which non-basic variable should be firstly considered as an entry variable • It makes use of stepping-stone to get its answer of next iteration • How it works? (to p31)

  31. Procedure (MODI) Step 0: let ui, v , cij variables represent rows, columns, and cost in the transportation tableau, respectively Step 1: (a) form a set of equations that uses to represent all basic variables ui + vj = cij (b) solve variables by assign one variable = 0 Step2: (a) form a set of equations use to represent non-basic variable (or empty cell) as such cij – ui – vj = kij (b) solve variables by using step 1b information Step 3: Select the cell that has the most –ve value in 2b Step 4: Use stepping-stone method to allocate resource to cell in step 3 Step 5: Repeat the above steps until all cells in 2a has no negative value (to p24) Example (to p32)

  32. MODI Consider to this initial tableau: Step 0: let ui, v , cij variables represent rows, columns, and cost in the transportation tableau, respectively (to p33)

  33. Step 0 Step 1: (a) form a set of equations that uses to represent all basic variables ui + vj = cij (to p34) C3A

  34. ui + vj = cij (to p35) (b) solve variables by assign one variable = 0

  35. Set one variable = 0 Because we added an non-basic variable (to p36) Step2: (a & b)

  36. Step2: (a & b) Note this may look difficult and complicated, however, we can add these V=values into the above tableau as well (to p37)

  37. Step2: (a & b), alternative -1 -1 +2 +5 (to p38) 6-0-7 Step 3: Select the cell that has the most –ve value in 2b

  38. Step3 -1 -1 +2 +5 (to p39) Select either one, (Why?) These cells mean, introduce it will reduce the min z to -1 cost unit

  39. Step 4: Use stepping-stone method - + + - From here we have …. (to p40)

  40. Step 4: Use stepping-stone method (to p41) Step 5: we repeat steps 1-4 again for the above tableau, we have

  41. Step 5 (to p42)

  42. Step 5 cont All positives STOP (to p31)

  43. Important Notes • When start solving a transportation problem using algorithm, we need to ensure the following: • Alternative solution • Total demand ≠ total supply • Degeneracy • others (to p44) (to p45) (to p48) (to p52) (to p7)

  44. Alternative solution • When on the following k = 0 cij – ui – vj = kij Why? (to p43)

  45. Total demand ≠ total supply Note that, total demand=650, and total supply = 600 How to solve it? We need to add a dummy row and assign o cost to each cell as such .. (to p46)

  46. Dd≠ss Extra row, since Demand > supply Other example …… (to p47)

  47. Dd≠ss Extra column is added (to p43)

  48. Degeneracy (to p49) Example ….. ie, basic variables in the tableau

  49. Degeneracy • m rows + n column – 1 = the number of cells with allocations • + 3 - 1 = 5 • It satisfied. • If failed? …… considering ….. (five basic variables, and above has 5 as well!) (to p50)

  50. Degeneracy (note above has only 4 basic variable only!) If not matched, then we select an non-basic variable with least cheapest cost and considered it as a new basic variable with assigned 0 allocation to it (to p51)